The area A between the graph of the function:
g(t) = 4 - (4/t^2)
and the t-axis over the interval [1, x] is:
A(x) = ∫[1, x] (4 - (4/t^2)) dt
a) Find the horizontal asymptote of the graph g.
I believe the horizontal asymptote of graph g is g(t) = 4.
b) Integrate to find A as a function of x. Does the graph of A have a horizontal asymptote? Explain.
I went ahead and computed the integral which I got
4t + (4/t) + C
But am I writing this correctly? I'm not exactly sure how question b wants me to write the answer out. Also how would I find if the graph of A have a horizontal asymptote?
Any help is greatly appreciated!
You have the right idea, but you are working a definite integral, so there is no constant C. Instead, evaluate A(t) at the interval endpoints:
A(x) = (4t + 4/t)[1,x]
= (4x + 4/x)-(4*1 + 4/1)
= 4x + 4/x - 8
a) To find the horizontal asymptote of the graph g, we need to determine the behavior of the function as t approaches infinity or negative infinity.
As t approaches infinity, the term 4/t^2 becomes smaller and smaller, approaching zero. Therefore, g(t) approaches 4 as t approaches infinity. Hence, the horizontal asymptote of the graph g is g(t) = 4.
b) To find A as a function of x, we can simply integrate the function g(t) with respect to t from 1 to x:
A(x) = ∫[1, x] (4 - (4/t^2)) dt
To integrate, we can rewrite the original function as:
A(x) = ∫[1, x] (4t^(-2) - 4) dt
Now, let's integrate each term separately. The integral of 4t^(-2) with respect to t is:
∫(4t^(-2)) dt = 4∫(1/t^2) dt
Using the power rule for integration, the integral of 1/t^2 is:
= -4/t
Now, let's integrate the constant term -4:
∫(-4) dt = -4t
Putting it all together, we have:
A(x) = 4∫(1/t^2) dt - 4t
= -4/t - 4t + C
So, the correct integral result is A(x) = -4/t - 4t + C. Note that the constant of integration, C, can vary.
To determine if the graph of A has a horizontal asymptote, we need to examine the behavior of A(x) as x approaches infinity or negative infinity.
As x approaches infinity, the term -4/t becomes smaller and smaller, approaching zero. The term -4t also becomes larger and larger, but there is no fixed value it approaches. Therefore, the graph of A does not have a horizontal asymptote.
I hope this helps! Let me know if you have any further questions.
a) To find the horizontal asymptote of the graph of g(t), we analyze the behavior of the function as t approaches positive or negative infinity.
For large positive values of t, the term 4/t^2 approaches zero, since the denominator becomes very large. Therefore, g(t) approaches 4 as t approaches positive infinity.
Similarly, for large negative values of t, the term 4/t^2 also approaches zero, since the denominator becomes very large. Therefore, g(t) approaches 4 as t approaches negative infinity.
Hence, the horizontal asymptote of the graph g is g(t) = 4.
b) Given the integral A(x) = ∫[1, x] (4 - (4/t^2)) dt, let's evaluate it. We need to compute the antiderivative of (4 - (4/t^2)) with respect to t.
∫(4 - (4/t^2)) dt = ∫(4 - 4t^(-2)) dt
To integrate this expression, we split the integral into two parts:
∫4 dt - ∫4t^(-2) dt
The antiderivative of 4 with respect to t is 4t, and the antiderivative of t^(-2) is -t^(-1) = -1/t.
∫4 dt - ∫4t^(-2) dt = 4t - (-4/t) + C = 4t + (4/t) + C
So, you are correct in your computation of the integral.
As for representing A as a function of x, we can write it as:
A(x) = 4x + 4/x + C
The constant C represents the arbitrary constant of integration, which you might have omitted in your working. This constant accounts for the fact that the integral has an infinite number of possible functions that have the same derivative.
Finally, to determine if the graph of A has a horizontal asymptote, we need to analyze the behavior of A(x) as x approaches positive or negative infinity.
As x becomes larger and larger, both the terms 4x and 4/x dominate the expression. The term 4x grows without bound, while the term 4/x approaches zero. This implies that as x approaches positive infinity, A(x) also approaches positive infinity.
Similarly, as x becomes more and more negative, both the terms 4x and 4/x dominate the expression. The term 4x becomes more and more negative, while the term 4/x approaches zero. This implies that as x approaches negative infinity, A(x) also approaches negative infinity.
Therefore, the graph of A(x) does not have a horizontal asymptote, as it approaches positive or negative infinity as x approaches infinity.